Example of a Nonlinear Operator | Math Definition

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A nonlinear operator can be exemplified by the operator defined as \hat{B} \psi = \psi^2, which demonstrates nonlinearity since \hat{B} (\psi_1 + \psi_2) does not equal \hat{B}(\psi_1) + \hat{B}(\psi_2). Nonlinear operators can take various forms, including arbitrary functions like sin, x^2, and log(x), contrasting with linear operators that follow the form f(x) = ax. The mathematical definition of a nonlinear operator states that it does not satisfy the condition A(au + bv) = aA(u) + bA(v) for all vectors u and v. Understanding nonlinear operators is crucial as they present more complex problems compared to linear operators, which allow for simpler solutions. Overall, nonlinear operators are essential in various mathematical and physical contexts, highlighting their significance in advanced studies.
Repetit
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Can someone give me an example of a nonlinear operator? My textbooks always proves that some operator is a linear operator, but I don't think I really know what a nonlinear operator looks like.

One of my books defines an operator like \hat{B} \psi = \psi^2. I see that this is a nonlinear operator because:

\hat{B} (\psi_1 + \psi_2) = (\psi_1 + \psi_2)^2

...and this is different from \psi_1^2 + \psi_2^2 which you would get by letting the operator B act on each function. But how can you define an operator like this? What would the mathematical form of such an operator be?
 
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An operator that is not linear?? In other words, an operator, A, is nonlinear if and only if there exist some vectors, u and v, and some numbers, a and b, such that A(au+ bv) is NOT equal to aA(u)+ bB(v). That's a perfectly good "mathematical definition". I not sure what you mean by a "mathematical form". What is the "mathematical form" of a linear operator?

Notice that if A(v)= v2[/sub] (assuming a one dimensional vector space for which squaring makes sense) then A(au+ bv)= (au+ bv)2= a2u2+ 2abuv+ b2v2 which is NOT the same as aA(u)+ bA(v)= au2+ bv2.

A two-dimensional example might be A((x,y))= (x2,y2). Then A(a(x,y)+ b(p,q))= A((ax+bp,ay+ bq))= (ax+bp)2, (ay+ bq)2= (a2x2+2abxp+ b2p2,a2y2+ 2abyq+ b2q2) which is not the same as aA(x,y)+ bA(p,q)= a(x2,y2)+ b(p2,q2)= (ax2+ bp2,ay2+ bq2).

Linear problems have the nice property that you can "take them apart", solve the simpler parts, and put those back together to get a solution to the original problem. With "non-linear" problems you can't do that. Essentially, "Linear Algebra" is the study of linear problems and so you very seldom have anything to do with non-linear operators. That is a much harder study!
 
Repetit said:
Can someone give me an example of a nonlinear operator? My textbooks always proves that some operator is a linear operator, but I don't think I really know what a nonlinear operator looks like.

they look like absolutely anything that is not linear. They are just arbitrary functions between spaces. f(x)=ax for some a are the only linear operators from R to R, for example, any other function, such as sin, x^2, log(x) and all the functions you know and love are non-linear operators.

One of my books defines an operator like \hat{B} \psi = \psi^2. I see that this is a nonlinear operator because:

\hat{B} (\psi_1 + \psi_2) = (\psi_1 + \psi_2)^2

...and this is different from \psi_1^2 + \psi_2^2 which you would get by letting the operator B act on each function. But how can you define an operator like this?

You just did define it.
 
Thanks a lot for the quick reply! What I meant was, what is the mathematical form of the operator \hat{B} that I introduced? For example, the form of the momentum operator \hat{p} in quamtum mechanics is - i \hbar \frac{d}{dx}, and the form of the laplace operator is:

\frac{d^2}{dx^2} + \frac{d^2}{dy^2} + \frac{d^2}{dz^2}

But what is the form of the operator B that I introduced (I believe you called it A)? What would I write instead of the operator?
 
An operator is just a function. In this case, it sends one function to another function. For example, d/dx sends the function x^2 to the function 2x. But any well defined function works as an operator, even if we have not assigned a nice symbol like d/dx to it, and the operator you defined in your first post is perfectly valid.
 
Thanks to all of you, I understand it now! :-)
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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